Chapter 01

Medium

Algebraic Structures & Axioms

00 · Symbol Glossary

Every symbol you'll see in this chapter, named and explained. When a new one appears, refer back here.

$\ast$Star — generic binary operation

A placeholder symbol for "some operation," used when we want to talk about structure without committing to $+$ , $\times$ , or any specific rule. $a \ast b$ just means "combine $a$ and $b$ using whatever operation we've fixed."

$(S, \ast)$Algebraic structure

A set $S$ paired with an operation $\ast$ defined on it. The parentheses notation emphasizes that a structure is both the set and the rule together — $(\mathbb{Z}, +)$ and $(\mathbb{Z}, \times)$ are different structures on the same underlying set.

$\mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{C}$Integers, rationals, reals, complex numbers

The standard number systems, each containing the previous one: $\mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R} \subset \mathbb{C}$ . These will be our default examples until we build new structures from scratch.

$\mathbb{Z}_n$Integers mod n

The set of remainders left after dividing by $n$ : $\{0, 1, 2, \ldots, n-1\}$ , equipped with addition (and multiplication) "wrapping around" at $n$ . Built formally in Section 06 — it will reappear in every later chapter.

$a \equiv b \pmod n$a is congruent to b mod n

Reads " $a$ is congruent to $b$ modulo $n$ ." Means $n$ divides $a - b$ — equivalently, $a$ and $b$ leave the same remainder when divided by $n$ . Example: $17 \equiv 5 \pmod{12}$ , the arithmetic behind clock time.

$\sim$Tilde — a relation, often an equivalence relation

A generic symbol for "is related to." $a \sim b$ means $a$ and $b$ are related in whatever way we've defined — not necessarily equal, but grouped together.

$[a]$Equivalence class of a

The set of every element related to $a$ under $\sim$ : $[a] = \{x \in S \mid x \sim a\}$ . All elements in $[a]$ are considered "the same" for the purposes of the relation.

$S/\!\sim$Quotient set

The set of all equivalence classes of $S$ under $\sim$ — a new set whose elements are themselves sets. Read " $S$ mod tilde." This exact construction reappears as quotient groups and quotient rings in later chapters.

$e$Identity element

The "do nothing" element of a structure: combining $e$ with anything leaves that thing unchanged. Additive identity is usually written $0$ , multiplicative identity $1$ ; $e$ is the generic symbol used when the operation is unspecified.

$a^{-1}$Inverse of a

The element that "undoes" $a$ : combining $a$ with $a^{-1}$ produces the identity. Written $-a$ for addition, $a^{-1}$ or $1/a$ for multiplication.

$\forall,\ \exists,\ \exists!$"for all", "there exists", "there exists a unique"

Quantifiers from formal logic. $\exists!$ (exists-unique) is stronger than $\exists$ : it asserts existence and asserts there is only one such object — exactly the claim made by uniqueness theorems in Section 03.

$\iff$If and only if

Biconditional — both directions of implication hold. $P \iff Q$ means $P$ implies $Q$ and $Q$ implies $P$ ; they are logically equivalent statements.

$\mid$Divides

$m \mid n$ reads " $m$ divides $n$ ," meaning $n = mk$ for some integer $k$ — no remainder. Distinct from the vertical bar used in set-builder notation, but context always makes the meaning clear.

01 · What Is Algebra, Really?

High-school algebra is about solving equations with numbers. Abstract algebra asks a different question: what makes the arithmetic of numbers work — and where else does that same working show up?

Addition of integers, multiplication of nonzero reals, composition of functions, symmetries of a triangle, shuffles of a deck of cards — these look unrelated. But strip away the specific objects and look only at the rules the combining operation obeys, and the same small list of patterns keeps reappearing: closure, associativity, an identity, inverses. Abstract algebra studies those patterns directly, as objects in their own right, so that a theorem proved once (say, "identities are unique") applies instantly to every structure satisfying the pattern — numbers, symmetries, matrices, polynomials, all at once.

This chapter builds the vocabulary. Nothing here is a new discovery about numbers; it is the scaffolding that lets Chapter 02 onward talk about groups, and eventually rings and fields, with total precision.

02 · Binary Operations

Definition — Binary Operation

A binary operation on a set $S$ is a rule $\ast$ that takes any two elements of $S$ (in order) and produces a single output that must also land in $S$ :

\ast : S \times S \to S, \qquad (a, b) \mapsto a \ast b

$S \times S$ — the set of all ordered pairs $(a,b)$ with $a, b \in S$ .

$\mapsto$ — "maps to." The pair $(a,b)$ is sent to the single output $a \ast b$ .

The requirement that the output land back in $S$ is called closure, and it is not automatic — it is a real condition that can fail.

Example — Ordinary operations as binary operations

Addition on $\mathbb{Z}$ : $\ast = +$ , and $+ : \mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$ sends $(3,5)\mapsto 8$ . Closed — the sum of two integers is always an integer.

Composition of functions $f,g : \mathbb{R}\to\mathbb{R}$ : $\ast = \circ$ , and $(f,g)\mapsto f\circ g$ , the function $x \mapsto f(g(x))$ . Closed — composing two real functions gives another real function.

Common mistake — Assuming closure for free

Is subtraction a binary operation on the natural numbers $\mathbb{N} = \{0,1,2,\ldots\}$ ? Take $a=3$ , $b=7$ : $3 - 7 = -4 \notin \mathbb{N}$ . The output escapes the set. Subtraction is not a binary operation on $\mathbb{N}$ — it fails closure, even though it's a perfectly good operation on $\mathbb{Z}$ .

This is why "is this closed?" is always the first question to ask about a candidate structure, and why so many exercises in this subject are disguised closure checks.

Example — An operation given by a table

For a finite set, a binary operation can be given entirely by a table, with no formula at all. Let $S = \{r, s\}$ and define $\ast$ by:

$\ast$	$r$	$s$
$r$	$r$	$s$
$s$	$s$	$r$

Read row-then-column: $r \ast s = s$ (row $r$ , column $s$ ). Every entry is in $\{r,s\}$ , so this is closed — and it is a complete, legitimate binary operation even though it has no numerical meaning at all. This table happens to describe the two symmetries of a non-square rectangle: $r$ = "do nothing," $s$ = "flip." You will meet it again, formally, in Chapter 02.

03 · Associativity and Commutativity

Closure alone gives very little to work with. Two further properties, each independently optional, are what make an operation usable.

Definition — Associativity

An operation $\ast$ on $S$ is associative if, for all $a,b,c \in S$ :

(a \ast b) \ast c = a \ast (b \ast c)

In words: when combining three (or more) elements in a fixed order, it does not matter which adjacent pair you combine first. Parentheses become optional, and $a \ast b \ast c$ is unambiguous.

Common mistake — Subtraction is not associative

$(8 - 5) - 2 = 3 - 2 = 1$ , but $8 - (5-2) = 8 - 3 = 5$ . Different answers — subtraction on $\mathbb{Z}$ is not associative. This is exactly why $8-5-2$ is only unambiguous by convention (read left to right), not by any inherent property of subtraction.

Definition — Commutativity

An operation $\ast$ on $S$ is commutative if, for all $a,b \in S$ :

a \ast b = b \ast a

Order of the two inputs does not matter. Commutativity and associativity are logically independent — a structure can have either, both, or neither.

Example — Associative but not commutative

Matrix multiplication on $2\times 2$ real matrices is associative ( $(AB)C = A(BC)$ always) but not commutative in general. For instance:

\begin{pmatrix}1&1\\0&1\end{pmatrix}\begin{pmatrix}1&0\\1&1\end{pmatrix} = \begin{pmatrix}2&1\\1&1\end{pmatrix}, \qquad \begin{pmatrix}1&0\\1&1\end{pmatrix}\begin{pmatrix}1&1\\0&1\end{pmatrix} = \begin{pmatrix}1&1\\1&2\end{pmatrix}

Different results depending on order. This single example is the template for every noncommutative group you will meet later — symmetries, permutations, and matrices all behave this way, and it is precisely this failure of commutativity that makes group theory richer than arithmetic.

Function composition is generally noncommutative too

$(f\circ g)(x) = f(g(x))$ but $(g\circ f)(x)=g(f(x))$ — usually different functions. Try $f(x)=x+1$ , $g(x)=2x$ : $f(g(x)) = 2x+1$ , while $g(f(x))=2x+2$ . Order of composition matters, exactly as order of matrix multiplication does — not a coincidence, since matrices represent linear functions.

04 · Identity Elements and Inverses

Definition — Identity Element

Let $(S,\ast)$ be a structure. An element $e \in S$ is an identity for $\ast$ if:

e \ast a = a \ast e = a \qquad \text{for every } a \in S

Combining $e$ with anything changes nothing. Both orders must give back $a$ — this matters when $\ast$ is not commutative.

Example — Identities you already know

$(\mathbb{Z}, +)$ : identity is $0$ , since $a + 0 = 0 + a = a$ .

$(\mathbb{R}, \times)$ : identity is $1$ , since $a \times 1 = 1 \times a = a$ .

$(\{\text{functions } \mathbb{R}\to\mathbb{R}\}, \circ)$ : identity is $\mathrm{id}(x) = x$ , since $f \circ \mathrm{id} = \mathrm{id}\circ f = f$ for every $f$ .

Before defining inverses, it is worth proving something that looks obvious but is not free: a structure cannot have two different identities.

Theorem 1.1 — Uniqueness of the Identity

If $(S,\ast)$ has an identity, it has exactly one.

Proof

Suppose $e$ and $e'$ are both identities for $\ast$ on $S$ . Since $e$ is an identity, applying it to $e'$ gives:

e \ast e' = e'

Since $e'$ is also an identity, applying it to $e$ gives:

e \ast e' = e

The left-hand sides of both equations are literally the same expression, $e \ast e'$ , so the right-hand sides must be equal: $e' = e$ . Hence any two identities coincide — there is at most one.

Why this proof pattern matters

This "combine the two candidates and read the answer two ways" trick is the single most common proof technique in algebra. You will see it again to prove inverses are unique, homomorphic images of identities are identities, and much more. Learn to recognize the shape: two things are each assumed to have property $P$ ; combine them; each assumption gives a different-looking expression for the same combination; equate.

Definition — Inverse

Let $(S,\ast)$ have identity $e$ . An element $b \in S$ is an inverse of $a \in S$ if:

a \ast b = b \ast a = e

If such a $b$ exists, $a$ is called invertible, and $b$ is written $a^{-1}$ (or $-a$ when $\ast$ is addition).

Common mistake — Not every element needs an inverse

In $(\mathbb{Z}, \times)$ , identity is $1$ . Does $3$ have an inverse? We'd need $b$ with $3b = 1$ — but $b = 1/3 \notin \mathbb{Z}$ . $3$ has no multiplicative inverse in $\mathbb{Z}$ . Only $1$ and $-1$ are invertible in $(\mathbb{Z},\times)$ . Invertibility is a per-element property, not automatic just because an identity exists — this is precisely the gap between a monoid and a group, made formal in Section 05.

Theorem 1.2 — Uniqueness of Inverses (associative case)

If $\ast$ is associative and has identity $e$ , then every invertible $a \in S$ has exactly one inverse.

Proof

Suppose $b$ and $c$ are both inverses of $a$ : $a\ast b = b\ast a = e$ and $a\ast c = c\ast a = e$ . Compute $b \ast a \ast c$ two ways, using associativity to regroup freely.

Group the left two first: $(b \ast a) \ast c = e \ast c = c$ , using $b\ast a = e$ and then the identity law.

Group the right two first: $b \ast (a \ast c) = b \ast e = b$ , using $a \ast c = e$ and then the identity law.

Both computations evaluate the same three-fold product $b \ast a \ast c$ — associativity guarantees the grouping doesn't change the value — so the two results must agree: $c = b$ . Hence the inverse is unique.

Associativity was essential

Look closely: the proof needed to regroup $b \ast a \ast c$ , which is only legal because $\ast$ is associative. Without associativity, "the inverse" would not even be well-defined language — there could genuinely be several unrelated elements $b$ satisfying $a\ast b = b\ast a = e$ . This is why every structure from here on (semigroups, monoids, groups, rings) builds associativity in from the start.

05 · A Hierarchy of Structures

Closure, associativity, identity, and inverses are four independent ingredients. Naming which subset a structure satisfies gives a standard hierarchy — a preview of names that will be defined properly, one at a time, starting with groups in Chapter 02.

Structure	Closure	Associative	Identity	Inverses	Example
Magma	✓	—	—	—	$(\mathbb{N}, -)$ restricted to stay nonnegative, informally
Semigroup	✓	✓	—	—	$(\mathbb{N}, +)$ — no identity if $0\notin\mathbb{N}$ 's convention used
Monoid	✓	✓	✓	—	$(\mathbb{Z}, \times)$ — identity $1$ , but $3$ has no inverse
Group	✓	✓	✓	✓	$(\mathbb{Z}, +)$ — identity $0$ , inverse of $a$ is $-a$

Reading the table

Each row adds a requirement to the row above — a group is a monoid in which every element happens to be invertible. This is why $(\mathbb{Z},\times)$ stalls at "monoid": it has an identity ( $1$ ) but most elements lack inverses. Swapping to $(\mathbb{Q}\setminus\{0\}, \times)$ upgrades all the way to a group, since every nonzero rational has a reciprocal.

Example — Classifying $(\mathbb{Z}, +)$, $(\mathbb{Z}, \times)$, and $(2\mathbb{Z}, +)$

$(\mathbb{Z}, +)$ : closed, associative, identity $0$ , every $a$ has inverse $-a$ . Group.

$(\mathbb{Z}, \times)$ : closed, associative, identity $1$ , but $2$ has no inverse in $\mathbb{Z}$ . Monoid, not a group.

$(2\mathbb{Z}, +)$ — the even integers under addition: sum of two evens is even (closed), associative (inherited from $\mathbb{Z}$ ), identity $0 \in 2\mathbb{Z}$ , inverse of $2k$ is $-2k \in 2\mathbb{Z}$ . Group. Notice this group lives inside $(\mathbb{Z},+)$ — exactly the idea behind subgroups in Chapter 03.

06 · Equivalence Relations & Partitions

One more piece of machinery is needed before this chapter can build $\mathbb{Z}_n$ , and it will resurface, unchanged, when quotient groups and quotient rings are built in later chapters: the idea of declaring certain elements "the same" without them being literally equal.

Definition — Equivalence Relation

A relation $\sim$ on a set $S$ is an equivalence relation if for all $a,b,c \in S$ :

Reflexive. $a \sim a$ .

Symmetric. $a \sim b \implies b \sim a$ .

Transitive. $a \sim b$ and $b \sim c \implies a \sim c$ .

Example — Equality is the smallest equivalence relation

Ordinary equality " $=$ " on any set satisfies all three: $a=a$ ; $a=b\implies b=a$ ; $a=b, b=c\implies a=c$ . Equivalence relations generalize equality by relaxing which elements count as interchangeable, while keeping the same three structural guarantees.

Common mistake — "Divides" is not an equivalence relation

Define $a \sim b$ on $\mathbb{Z}^+$ by $a \mid b$ ( $a$ divides $b$ ). Reflexive: yes, $a \mid a$ . Transitive: yes. Symmetric: no — $2 \mid 4$ but $4 \nmid 2$ . Failing even one of the three conditions disqualifies a relation from being an equivalence relation, no matter how natural it looks.

Definition — Equivalence Class

For $a \in S$ , the equivalence class of $a$ is:

[a] = \{\, x \in S \mid x \sim a \,\}

the set of everything related to $a$ . Note $a \in [a]$ always, by reflexivity.

Theorem 1.3 — Equivalence Classes Partition the Set

If $\sim$ is an equivalence relation on $S$ , the equivalence classes are pairwise disjoint or identical, and their union is all of $S$ . Equivalently: every element of $S$ lies in exactly one equivalence class.

Proof

Union is everything. For any $a \in S$ , reflexivity gives $a \sim a$ , so $a \in [a]$ . Hence every element belongs to some class, and the union of all classes is $S$ .

Distinct classes don't overlap. Suppose $[a] \cap [b] \neq \varnothing$ — some $c$ satisfies $c \sim a$ and $c \sim b$ . By symmetry, $a \sim c$ . By transitivity applied to $a \sim c$ and $c \sim b$ , we get $a \sim b$ .

Now take any $x \in [a]$ , so $x \sim a$ . Since $a \sim b$ , transitivity gives $x \sim b$ , so $x \in [b]$ . This shows $[a] \subseteq [b]$ . The symmetric argument (swap the roles of $a$ and $b$ ) gives $[b] \subseteq [a]$ . Hence $[a] = [b]$ .

So two classes are either completely disjoint or exactly equal — never partially overlapping. Combined with the first part, every element sits in precisely one class, i.e. the classes partition $S$ .

Why this theorem is worth remembering by name

This exact three-line argument — "shared element forces the classes to merge" — is the mechanism behind cosets partitioning a group in Chapter 04, and behind the fact that quotient structures (quotient groups, quotient rings) are always built from a genuine partition, never from an ambiguous or overlapping grouping. Once you've internalized this proof, several later theorems become "the same proof, different notation."

07 · Modular Arithmetic — A Running Example

Everything above becomes concrete in a single construction that will recur in every remaining chapter of this subject.

Definition — Congruence mod n

Fix a positive integer $n$ . For $a, b \in \mathbb{Z}$ , say $a$ is congruent to $b$ modulo $n$ , written $a \equiv b \pmod n$ , if:

n \mid (a-b)

Equivalently, $a$ and $b$ leave the same remainder when divided by $n$ .

Theorem 1.4 — Congruence mod n is an Equivalence Relation

For fixed $n$ , the relation $\equiv \pmod n$ is an equivalence relation on $\mathbb{Z}$ .

Proof

Reflexive. $a - a = 0 = n\cdot 0$ , so $n \mid (a-a)$ ; hence $a \equiv a \pmod n$ .

Symmetric. If $a \equiv b \pmod n$ , then $a - b = nk$ for some integer $k$ . Then $b - a = n(-k)$ , and $-k \in \mathbb{Z}$ , so $n \mid (b-a)$ ; hence $b \equiv a \pmod n$ .

Transitive. If $a \equiv b \pmod n$ and $b \equiv c \pmod n$ , then $a-b = nk$ and $b-c=nj$ for integers $k,j$ . Adding: $(a-b)+(b-c) = a - c = n(k+j)$ , and $k+j \in \mathbb{Z}$ , so $n \mid (a-c)$ ; hence $a \equiv c \pmod n$ .

By Theorem 1.3, congruence mod $n$ partitions $\mathbb{Z}$ into equivalence classes. This partition is the object we actually want.

Definition — The Set $\mathbb{Z}_n$

The set of equivalence classes of $\mathbb{Z}$ under $\equiv \pmod n$ is denoted $\mathbb{Z}_n$ :

\mathbb{Z}_n = \{\, [0], [1], [2], \ldots, [n-1] \,\}

There are exactly $n$ classes, one per possible remainder, since every integer is congruent to its remainder on division by $n$ , and two different remainders in $\{0,\ldots,n-1\}$ are never congruent to each other (their difference is smaller than $n$ in absolute value and nonzero, so $n$ cannot divide it).

Step-by-step — Sorting integers into classes of $\mathbb{Z}_4$

Set $n=4$ . The possible remainders on division by 4 are

0,1,2,3

— exactly 4 classes.

Locate $-1$ :

-1 = 4(-1) + 3

, remainder

3

. So

-1 \in [3]

Locate $9$ :

9 = 4(2)+1

, remainder

1

. So

9 \in [1]

Locate $16$ :

16=4(4)+0

, remainder

0

. So

16 \in [0]

Conclusion:

\mathbb{Z}_4 = \{[0],[1],[2],[3]\}

, and every integer, no matter how large or negative, lands in exactly one of these four classes.

We now want to add and multiply classes, not just integers. The natural definition is $[a]+[b] := [a+b]$ — but this is only legitimate if the answer does not depend on which representative of each class we happened to pick.

Theorem 1.5 — Addition and Multiplication on $\mathbb{Z}_n$ Are Well-Defined

If $a \equiv a' \pmod n$ and $b \equiv b' \pmod n$ , then:

a+b \equiv a'+b' \pmod n \qquad \text{and} \qquad ab \equiv a'b' \pmod n

Consequently, $[a]+[b] := [a+b]$ and $[a][b] := [ab]$ are legitimate operations on $\mathbb{Z}_n$ — the result does not depend on the choice of representative.

Proof

By hypothesis, $a - a' = nk$ and $b - b' = nj$ for some integers $k,j$ .

Addition. $(a+b) - (a'+b') = (a-a') + (b-b') = nk + nj = n(k+j)$ , and $k+j\in\mathbb{Z}$ . So $n \mid \big((a+b)-(a'+b')\big)$ , giving $a+b \equiv a'+b' \pmod n$ .

Multiplication. Write $a = a' + nk$ and $b = b'+nj$ . Then:

ab = (a'+nk)(b'+nj) = a'b' + a'nj + nkb' + n^2kj = a'b' + n(a'j + kb' + nkj)

The quantity $a'j+kb'+nkj$ is an integer, so $ab - a'b' = n(a'j+kb'+nkj)$ is a multiple of $n$ . Hence $ab \equiv a'b' \pmod n$ .

Since the congruence class of the sum/product depends only on the classes of $a$ and $b$ — not on which representative was used to compute it — the operations $[a]+[b]:=[a+b]$ and $[a][b]:=[ab]$ are unambiguous rules on $\mathbb{Z}_n$ .

Well-definedness is a recurring checkpoint

Whenever a later chapter defines an operation on a quotient object — a quotient group $G/N$ , a quotient ring $R/I$ — this exact question resurfaces: does the formula depend on which representative you picked? The proof pattern here (write the difference as a multiple of $n$ , expand, factor $n$ back out) is the template used again, essentially unchanged, in Chapter 05 and Chapter 12.

Example — The addition table of $\mathbb{Z}_4$

Writing classes as bare numbers $0,1,2,3$ for brevity (each really means $[0],[1],[2],[3]$ ):

$+$	0	1	2	3
0	0	1	2	3
1	1	2	3	0
2	2	3	0	1
3	3	0	1	2

Every row and column contains each element exactly once, and the table is symmetric across the diagonal (commutative). Identity is $0$ (its row/column reproduces the header). Every element has an inverse: $1+3=0$ , $2+2=0$ , $3+1=0$ . This is $(\mathbb{Z}_4, +)$ satisfying every group axiom you will meet formally in Chapter 02 — it is, in fact, the very first genuinely new group in this course, distinct from anything built purely from $\mathbb{Z}$ , $\mathbb{Q}$ , or $\mathbb{R}$ .

Common mistake — $(\mathbb{Z}_n, \times)$ is usually only a monoid

Look at the multiplication table restricted to $\mathbb{Z}_4$ : $2 \times 2 = [4] = [0]$ . Two nonzero classes multiplied together gave the "zero" class — something that never happens in $\mathbb{Z}$ , $\mathbb{Q}$ , or $\mathbb{R}$ . Consequently $2$ has no multiplicative inverse in $\mathbb{Z}_4$ : there is no $x$ with $2x \equiv 1 \pmod 4$ (checking $x=0,1,2,3$ gives products $0,2,0,2$ — never $1$ ). So $(\mathbb{Z}_4,\times)$ is a monoid (identity $1$ ) but not a group. This "zero times nonzero equals zero" failure, called a zero divisor, becomes an entire topic of its own — it's precisely why Chapter 15 needs to distinguish domains from general rings.

08 · Exercises

EXERCISE 1.1

Check the three conditions in order: does the output always land back in $S$ ? Try a potential counterexample with small numbers before assuming yes.

Let $a,b \in 3\mathbb{Z}$ , so $a=3m$ , $b=3k$ for integers $m,k$ . Then $a+b = 3(m+k)$ , and $m+k \in \mathbb{Z}$ , so $a+b \in 3\mathbb{Z}$ . Closed.

Associativity and commutativity are inherited from $(\mathbb{Z},+)$ since $3\mathbb{Z} \subseteq \mathbb{Z}$ and the addition rule is the same operation, just restricted.

Identity: $0 = 3(0) \in 3\mathbb{Z}$ , and $a+0=a$ . Identity exists.

Inverses: for $a=3m$ , $-a = 3(-m) \in 3\mathbb{Z}$ , and $a+(-a)=0$ . Every element has an inverse.

All four properties hold: $(3\mathbb{Z}, +)$ is a group.

Is $(3\mathbb{Z}, +)$ — the multiples of 3 under addition — a group? Check closure, associativity, identity, and inverses explicitly.

EXERCISE 1.2

Apply Theorem 1.2's proof directly to this specific structure: what is $e$ here, and can you exhibit two candidate inverses that must be shown equal?

In $(\mathbb{R}\setminus\{0\}, \times)$ , the identity is $e=1$ . Suppose $b$ and $c$ both satisfy $ab=ba=1$ and $ac=ca=1$ . Then, exactly as in Theorem 1.2's proof:

$(b\cdot a)\cdot c = 1 \cdot c = c, \qquad b \cdot (a \cdot c) = b \cdot 1 = b$

Multiplication of real numbers is associative, so both expressions equal $b\cdot a\cdot c$ , forcing $b=c$ . This confirms directly what Theorem 1.2 already guarantees in general: real multiplicative inverses ( $1/a$ ) are unique.

Without quoting Theorem 1.2 by name, reprove directly that inverses in $(\mathbb{R}\setminus\{0\}, \times)$ are unique.

EXERCISE 1.3

Reflexive and transitive are usually easy. For symmetric, try to find two integers $a,b$ where $a \le b$ but not $b \le a$ .

Define $a \sim b$ to mean $a \le b$ on $\mathbb{Z}$ . Reflexive: $a \le a$ , true. Transitive: $a\le b, b\le c \implies a\le c$ , true. Symmetric: would need $a\le b \implies b \le a$ for all $a,b$ — false, e.g. $2 \le 5$ but $5 \not\le 2$ .

Not an equivalence relation — it fails symmetry. ( $\le$ is instead an example of a partial order, a different structure entirely.)

Is $a \sim b \iff a \le b$ an equivalence relation on $\mathbb{Z}$ ? Justify by checking all three conditions.

EXERCISE 1.4

Find the remainder of each number after division by 5, then group numbers sharing the same remainder.

Remainders mod 5: $12 = 5(2)+2 \to [2]$ . $-3 = 5(-1)+2 \to [2]$ . $7 = 5(1)+2 \to [2]$ . $20 = 5(4)+0 \to [0]$ . $2 = 5(0)+2 \to [2]$ .

So $12, -3, 7, 2$ all land in $[2]$ , and $20$ alone lands in $[0]$ . Two classes appear: $[2] = \{12,-3,7,2\}$ and $[0]=\{20\}$ , consistent with Theorem 1.3 — every element falls in exactly one class, and elements in the same class really do share a remainder.

Sort $12, -3, 7, 20, 2$ into equivalence classes under $\equiv \pmod 5$ .

EXERCISE 1.5

Build the full $5\times 5$ multiplication table for $\mathbb{Z}_5\setminus\{0\}$ mentally or on paper, and check each row for whether $1$ appears.

Multiplying every nonzero pair mod 5: $1\cdot4=4$ , $2\cdot3=6\equiv1$ , $2\cdot4=8\equiv3$ , $3\cdot4=12\equiv2$ , and each element times itself: $2\cdot2=4$ , $3\cdot3=9\equiv4$ , $4\cdot4=16\equiv1$ .

Checking inverses: $1\cdot1=1$ ( $1^{-1}=1$ ); $2\cdot3=6\equiv1$ ( $2^{-1}=3$ ); $3\cdot2=6\equiv1$ ( $3^{-1}=2$ ); $4\cdot4=16\equiv1$ ( $4^{-1}=4$ ).

Every nonzero element of $\mathbb{Z}_5$ has a multiplicative inverse — unlike $\mathbb{Z}_4$ . The difference: $5$ is prime, so no nonzero remainder shares a common factor with $5$ , and (as Chapter 15 will make precise) that is exactly the condition for multiplicative inverses to exist mod $n$ .

In $\mathbb{Z}_5 = \{0,1,2,3,4\}$ , does every nonzero element have a multiplicative inverse? Find each one, or explain why one fails to exist.

EXERCISE 1.6

Mimic the two-column proof of Theorem 1.1 exactly, but swap "identity" for "absorbing element": what equation does $z\ast a = z$ give you, twice, in two different ways?

Call $z$ absorbing if $z \ast a = a \ast z = z$ for every $a \in S$ . Suppose $z$ and $z'$ are both absorbing. Apply $z$ 's property to $a = z'$ : $z \ast z' = z$ . Apply $z'$ 's property to $a=z$ : $z \ast z' = z'$ . The left-hand sides are the same expression, so the right-hand sides agree: $z = z'$ .

An absorbing element, if one exists, is unique — by the identical proof strategy as Theorem 1.1, with the roles of the defining equation swapped.

An element $z \in S$ is called absorbing for $\ast$ if $z \ast a = a \ast z = z$ for every $a \in S$ (e.g. $0$ is absorbing for multiplication: $0 \times a = 0$ ). Prove that if a structure has an absorbing element, it has only one.

09 · Chapter Summary

Concept	Statement
Binary operation	$\ast: S\times S\to S$ ; must satisfy closure
Associative	$(a\ast b)\ast c = a\ast(b\ast c)$
Commutative	$a\ast b = b\ast a$
Identity	$e\ast a = a\ast e = a$ ; unique when it exists (Thm 1.1)
Inverse	$a\ast a^{-1}=a^{-1}\ast a=e$ ; unique under associativity (Thm 1.2)
Hierarchy	Magma $\subset$ Semigroup $\subset$ Monoid $\subset$ Group
Equivalence relation	Reflexive, symmetric, transitive
Equivalence classes	Partition the set (Thm 1.3)
$\mathbb{Z}_n$	Equivalence classes of $\mathbb{Z}$ under $\equiv \pmod n$
Well-definedness on $\mathbb{Z}_n$	$+, \times$ don't depend on choice of representative (Thm 1.5)

Next: Chapter 02 — Groups formalizes the "closure + associative + identity + inverses" row of the hierarchy table above as a single object of study, and starts building the standard library of examples: symmetric groups, dihedral groups, and $(\mathbb{Z}_n, +)$ revisited from a group-theoretic point of view.